Phy5645/Hydrogen Atom WKB: Difference between revisions

Latest revision as of 13:45, 18 January 2014

The WKB approximation is given by

$\int _{r_{1}}^{r_{2}}p(r)\,dr=(n+{\tfrac {1}{2}})\pi \hbar ,$

where

$p(r)={\sqrt {2m(E-V_{\text{eff}}(r))}}={\sqrt {2m\left(E+{\frac {e^{2}}{r}}-{\frac {\hbar ^{2}(l+{\tfrac {1}{2}})^{2}}{2mr^{2}}}\right)}}.$

We may rewrite the above as

${\sqrt {2mE}}\int _{r_{1}}^{r_{2}}{\sqrt {1-{\frac {\hbar ^{2}(l+{\tfrac {1}{2}})^{2}}{2mEr^{2}}}+{\frac {e^{2}}{Er}}}}\,dr=(n+{\tfrac {1}{2}})\pi \hbar ,$

or, making the substitution,

$T=-{\frac {\hbar ^{2}(l+{\tfrac {1}{2}})}{2mE}}$ and $V={\frac {e^{2}}{E}},$

${\sqrt {2mE}}\int _{r_{1}}^{r_{2}}{\sqrt {1-{\frac {V}{r}}+{\frac {T}{r^{2}}}}}\,dr=(n+{\tfrac {1}{2}})\pi \hbar .$

Using the fact that $r^{2}-Vr+T=(r_{1}-r)(r_{2}-r)\!$ and that

$\int _{r_{1}}^{r_{2}}{\sqrt {\frac {(x-r_{1})(x-r_{2})}{x^{2}}}}\,dx={\frac {\pi }{2}}({\sqrt {r_{2}}}-{\sqrt {r_{1}}})^{2},$

we obtain

${\frac {\pi }{2}}{\sqrt {2mE}}({\sqrt {r_{2}}}-{\sqrt {r_{1}}})^{2}=(n+{\tfrac {1}{2}})\pi \hbar .$

We now observe that $r^{2}-Vr+T=(r_{1}-r)(r_{2}-r)=r^{2}-(r_{1}+r_{2})r+r_{1}r_{2},\!$ so that

$V=r_{1}+r_{2}\!$ and $T=r_{1}r_{2}.\!$

We thus obtain

${\frac {\pi }{2}}{\sqrt {2mE}}(V-2{\sqrt {T}})=(n+{\tfrac {1}{2}})\pi \hbar ,$

or

${\sqrt {2mE}}\left({-{\frac {e^{2}}{E}}-2{\sqrt {-{\frac {\hbar ^{2}(l+{\tfrac {1}{2}})^{2}}{2mE}}}}}\right)=(n+{\tfrac {1}{2}})\pi \hbar .$

This equation simplifies to

$-e^{2}{\sqrt {-{\frac {2m}{E}}}}-(2\ell +1)\hbar =(2n+1)\hbar .$

We may now easily solve for $E,\!$ obtaining

$E=-{\frac {me^{4}}{2\hbar ^{2}(n+\ell +1)^{2}}}.$

Note that this is exactly the spectrum that we would obtain from an exact solution of the Coulomb problem.

Back to WKB in Spherical Coordinates

@@ Line 1: / Line 1: @@
-Use WKB approximation to estimate energy spectrum for Hydrogen atom.
+The WKB approximation is given by
-Hint: <math>\text{use the relation let   r}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)</math>
-</math>
-----
-----
-The approximation is:
-<math>\int {P(r)} dr=(n+\frac{1}{2})\pi \hbar</math>
+<math>\int_{r_1}^{r_2} p(r)\,dr=(n+\tfrac{1}{2})\pi \hbar,</math>
-<math>\text{P(r)=}\sqrt {2m(E-V(r))} =\sqrt {2m(E-\left ({\frac{\hbar ^{2}l(l+1)}{2mr^{2}}-\frac{e^{2}}{r}} \right )} )</math>
+where
-<math>\int\limits_{r1}^{r2} {\sqrt {2m(E-\frac{\hbar ^{2}l(l+1)}{2mr^{2}}+\frac{e^{2}}{r})} }dr=(n+\frac{1}{2})\pi \hbar </math>  where r1 and r2 are turning points in this case.
+<math>p(r)=\sqrt {2m(E-V_{\text{eff}}(r))} =\sqrt {2m\left (E+\frac{e^{2}}{r}-{\frac{\hbar ^{2}(l+\tfrac{1}{2})^2}{2mr^{2}}}\right )}.</math>
-<math>\sqrt {2mE} \int\limits_{r1}^{r2} {(1-}\frac{\hbar ^{2}l(l+1)}{2mr^{2}E}+\frac{e^{2}}{Er})^{1/2}dr=(n+\frac{1}{2})\pi \hbar </math>
+We may rewrite the above as
-if we do this substitution:
-<math>\text{let T=}-\frac{\hbar ^{2}l(l+1)}{2mE}\text{   and V=}-\frac{e^{2}}{r}\text{    }</math>
-<math>\sqrt {2mE} \int\limits_{r1}^{r2} {(1+\frac{T}{r^{2}}}+\frac{V}{r})^{1/2}dr=(n+\frac{1}{2})\pi \hbar \text{ }</math>
+<math>\sqrt{2mE}\int_{r_1}^{r_2}\sqrt{1-\frac{\hbar^{2}(l+\tfrac{1}{2})^2}{2mEr^{2}}+\frac{e^{2}}{Er}}\,dr=(n+\tfrac{1}{2})\pi \hbar,</math>
-<math>\text{the relation  r}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)</math>
+or, making the substitution,
-<math>\sqrt {2mE} \int\limits_{r1}^{r2} {\left ({\frac{(r_{1}-r)(r_{2}-r)}{r^{2}}} \right )^{1/2}dr=(n+\frac{1}{2})\pi \hbar }</math>
+<math>T=-\frac{\hbar^{2}(l+\tfrac{1}{2})}{2mE}</math> and <math>V=\frac{e^{2}}{E},</math>
-<math>\text{ use definition  let  }\int\limits_{r1}^{r2} {\left ({\frac{(x-a)(x-b)}{x^{2}}} \right )^{1/2}dx}=\frac{\pi }{2}(\sqrt {b} -\sqrt {a} )^{2}</math>
+<math>\sqrt{2mE}\int_{r_1}^{r_2}\sqrt{1-\frac{V}{r}+\frac{T}{r^{2}}}\,dr=(n+\tfrac{1}{2})\pi \hbar.</math>
-<math>\sqrt {2mE} *\frac{\pi }{2}*(\sqrt {r_{2}} -\sqrt {r_{1}} )^{2}=(n+\frac{1}{2})\pi \hbar </math>
+Using the fact that <math>r^{2}-Vr+T=(r_{1}-r)(r_{2}-r)\!</math> and that
-<math>\sqrt {2mE} *\frac{\pi }{2}*(r_{2}+r_{1}-2\sqrt {r_{1}r_{2}} )=(n+\frac{1}{2})\pi \hbar </math>
+<math>\int_{r_1}^{r_2}\sqrt{{\frac{(x-r_1)(x-r_2)}{x^{2}}}}\,dx=\frac{\pi }{2}(\sqrt {r_2} -\sqrt {r_1} )^{2},</math>
-<math>\text{let r}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)=r^{2}-(r_{1}+r_{2})+r_{1}r_{2}</math>
+we obtain
-<math>\text{so V=}(r_{1}+r_{2})\text{   and T=}r_{1}r_{2}</math>
+<math>\frac{\pi}{2}\sqrt {2mE}(\sqrt {r_{2}} -\sqrt {r_{1}} )^{2}=(n+\tfrac{1}{2})\pi \hbar.</math>
-<math>\sqrt {2mE} *\frac{\pi }{2}*(V-2\sqrt {T} )=(n+\frac{1}{2})\pi \hbar </math>
+We now observe that <math>r^{2}-Vr+T=(r_{1}-r)(r_{2}-r)=r^{2}-(r_{1}+r_{2})r+r_{1}r_{2},\!</math> so that
-<math>\sqrt {2mE} \left ({-\frac{e^{2}}{E}-2\sqrt {-\frac{\hbar ^{2}l(l+1)}{2mE}} } \right )=(n+\frac{1}{2})\pi \hbar </math>
+<math>V=r_{1}+r_{2}\!</math> and <math>T=r_{1}r_{2}.\!</math>
-<math>-e^{2}\sqrt {\frac{2m}{E}} -2\sqrt {\hbar ^{2}l(l+1)} =2\hbar (n+\frac{1}{2})</math>
+We thus obtain
-<math>\text{ }2\hbar (n+\frac{1}{2})+2\hbar \sqrt {l(l+1)} =e^{2}\sqrt {\frac{2m}{-E}} </math>
+<math>\frac{\pi }{2}\sqrt {2mE}(V-2\sqrt {T} )=(n+\tfrac{1}{2})\pi \hbar,</math>
-<math>\frac{4\hbar ^{2}\left ({n+\frac{1}{2}+\sqrt {l(l+1)} } \right )^{2}}{2me^{4}}=\frac{1}{-E}\text{ }</math>
+or
-<math>\text{E=}\frac{-me^{4}}{2\hbar ^{2}\left ({n+\frac{1}{2}+\sqrt {l(l+1)} } \right )^{2}}</math>
+<math>\sqrt {2mE} \left ({-\frac{e^{2}}{E}-2\sqrt {-\frac{\hbar ^{2}(l+\tfrac{1}{2})^2}{2mE}} } \right )=(n+\tfrac{1}{2})\pi \hbar.</math>
+This equation simplifies to
+<math>-e^{2}\sqrt {-\frac{2m}{E}} -(2\ell+1)\hbar =(2n+1)\hbar.</math>
+We may now easily solve for <math>E,\!</math> obtaining
+<math>E=-\frac{me^{4}}{2\hbar ^{2}(n+\ell+1)^{2}}.</math>
+Note that this is exactly the spectrum that we would obtain from an exact solution of the Coulomb problem.
+Back to [[WKB in Spherical Coordinates#Problem|WKB in Spherical Coordinates]]

Phy5645/Hydrogen Atom WKB: Difference between revisions

Latest revision as of 13:45, 18 January 2014

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